Zhero wrote: Let $a,b,c$ be positive reals. Prove that \[ \frac{(a-b)(a-c)}{2a^2 + (b+c)^2} + \frac{(b-c)(b-a)}{2b^2 + (c+a)^2} + \frac{(c-a)(c-b)}{2c^2 + (a+b)^2} \geq 0. \] Let $c=\min\{a,b,c\}$. Hence, $\sum_{cyc} \frac{(a-b)(a-c)}{2a^2 + (b+c)^2}\geq\frac{(a-b)(a-c)}{2a^2 + (b+c)^2} + \frac{(b-c)(b-a)}{2b^2 + (c+a)^2}=$ $=\frac{(a-b)^2(a^2-ab+b^2+3c(a+b)-c^2)}{(2a^2 + (b+c)^2)(2b^2 + (c+a)^2)}\geq0$. The following inequality is also true. Let $a$, $b$ and $c$ are positives. Prove that: \[ \frac{(a-b)(a-c)}{3a^2 + (b+c)^2} + \frac{(b-c)(b-a)}{3b^2 + (c+a)^2} + \frac{(c-a)(c-b)}{3c^2 + (a+b)^2} \geq 0 \]

We can assume WLOG that $ a \geq b \geq c $. Now the Vornicu-Schur inequality states that if $ x, y, z $ are nonnegative reals satisfying $ x + z \geq y $ then $ x(a - b)(a - c) + y(b - a)(b - c) + z(c - a)(c - b) \geq 0 $. So it suffices to show that $ \frac{1}{2a^2 + (b + c)^2} + \frac{1}{2c^2 + (a + b)^2} \geq \frac{1}{2b^2 + (c + a)^2} $. Now, since the inequalities $ (b + c)^2 \leq 2b^2 + 2c^2 $ and $ (a + b)^2 \leq 2a^2 + 2b^2 $ hold, we have that $ \frac{1}{2a^2 + (b + c)^2} + \frac{1}{2c^2 + (a + b)^2} \geq \frac{1}{a^2 + b^2 + c^2} \geq \frac{1}{2b^2 + (c + a)^2} $ as desired, so we are done.

arqady wrote: Let $a$, $b$ and $c$ are positives. Prove that: \[ \frac{(a-b)(a-c)}{3a^2 + (b+c)^2} + \frac{(b-c)(b-a)}{3b^2 + (c+a)^2} + \frac{(c-a)(c-b)}{3c^2 + (a+b)^2} \geq 0 \] The same idea to your solution~ Let $ c=\min\{a,b,c\} $ then we would get: $ LHS \geq \frac{(a-b)(a-c)}{3a^2 + (b+c)^2} + \frac{(b-c)(b-a)}{3b^2 + (c+a)^2} $ $ =\frac{(a-b)^2((a-b)^2+(4c(a+b)-c^2))}{(3a^2 + (b+c)^2)(3b^2 + (c+a)^2)} \geq 0 $

Zhero wrote: Let $a,b,c$ be positive reals. Prove that \[ \frac{(a-b)(a-c)}{2a^2 + (b+c)^2} + \frac{(b-c)(b-a)}{2b^2 + (c+a)^2} + \frac{(c-a)(c-b)}{2c^2 + (a+b)^2} \geq 0. \] Calvin Deng. For the original and the one by Arqady, you can use Vornicu-Schur inequality. How about this one? Let $a,b,c$ be positive reals. Prove that \[ \sum_{cyc}\frac{(a-2b)(a-c)}{a^2 + (b+c)^2} \geq 0. \]

BSJL wrote: The same idea to your solution~ Yes! My method is much stronger than Vornicu-Schur.

My_dad wrote: How about this one? Let $a,b,c$ be positive reals. Prove that \[ \sum_{cyc}\frac{(a-2b)(a-c)}{a^2 + (b+c)^2} \geq 0. \] It's equivalent to $\sum_{cyc}(a^6+a^5b+a^5c-3a^4b^2-3a^4c^2+2a^3b^3+5a^4bc-6a^3b^2c-2a^3c^2b+4a^2b^2c^2)\geq0$. Since $\sum_{cyc}(a^3c^2b-a^2b^2c^2)\geq0$, it remains to prove that $\sum_{cyc}(a^6+a^5b+a^5c-3a^4b^2-3a^4c^2+2a^3b^3+5a^4bc-6a^3b^2c-6a^3c^2b+8a^2b^2c^2)\geq0$, which is easy by Schur and SOS: Let $a\geq b\geq c$. Hence, $\sum_{cyc}(a^6+a^5b+a^5c-3a^4b^2-3a^4c^2+2a^3b^3+5a^4bc-6a^3b^2c-6a^3c^2b+8a^2b^2c^2)=$ $\sum_{cyc}(a^6-a^5b-a^5c+a^4bc)+\sum_{cyc}(2a^5b+2a^5c-2a^4b^2-2a^4c^2-a^4b^2-a^4c^2+2a^3b^3)+$ $+2abc\sum_{cyc}(2a^3-a^2b-a^2c-2a^2b-2a^2c+4abc)\geq$ $\geq\sum_{cyc}ab(2a^2+ab+2b^2)(a-b)^2+2abc\sum_{cyc}(a+b-2c)(a-b)^2\geq$ $\geq ac(2a^2+ac+2c^2+2ab+2bc-4b^2)(a-c)^2+$ $+bc(2b^2+bc+2c^2+2ab+2ac-4a^2)(b-c)^2\geq$ $\geq ac(2a^2+2ab-4b^2)(b-c)^2+bc(2b^2+2ab-4a^2)(b-c)^2=$ $=2c(a+b)(a-b)^2(b-c)^2\geq0$.

Zhero wrote: Let $a,b,c$ be positive reals. Prove that \[ \frac{(a-b)(a-c)}{2a^2 + (b+c)^2} + \frac{(b-c)(b-a)}{2b^2 + (c+a)^2} + \frac{(c-a)(c-b)}{2c^2 + (a+b)^2} \geq 0. \] Calvin Deng. SOS-Schur one line Let $a=min\{a,b,c\}$ then $LHS=(\frac{1}{2a^2+(b+c)^2})(a-b)(a-c)+(b-c)^2(\frac{b^2+c^2-a^2+3a(b+c)-bc}{(2b^2+(c+a)^2)(2c^2+(a+b)^2)}) \geq 0$ Your sincerely CeuAzul

Zhero wrote: Let $a,b,c$ be positive reals. Prove that \[ \frac{(a-b)(a-c)}{2a^2 + (b+c)^2} + \frac{(b-c)(b-a)}{2b^2 + (c+a)^2} + \frac{(c-a)(c-b)}{2c^2 + (a+b)^2} \geq 0. \] Calvin Deng. The following inequality is also true. Let $a,b,c$ be positive reals. Prove that \[ \frac{(a-b)(a-c)}{2a^2 + (b+c)^2} + \frac{(b-c)(b-a)}{2b^2 + (c+a)^2} + \frac{(c-a)(c-b)}{2c^2 + (a+b)^2} \geq \frac{16}{35}\left(1-\frac{ab+bc+ca}{a^2+b^2+c^2}\right) \]

My_dad wrote: For the original and the one by Arqady, you can use Vornicu-Schur inequality. How about this one? Let $a,b,c$ be positive reals. Prove that \[ \sum_{cyc}\frac{(a-2b)(a-c)}{a^2 + (b+c)^2} \geq 0. \]

I think SOS can kill it . $(a-b)(a-c)=\frac{1}{2}((a-b)^2+(a-c)^2-(b-c)^2)$

Just use cs for the denominator we r done ! 2($b^2$+$c^2$)≥$(b+c)^2$ and expand the numerators simple

@above \[\frac{(a-b)(a-c)}{2a^2 + (b+c)^2} \ge \frac{(a-b)(a-c)}{2(a^2+b^2+c^2)}\]Is not always true, so your proof is not correct (what I mean is that this fact does not trivialize the problem). See post #3 for a proof using $2(b^2+c^2) \ge (b+c)^2$. @below, what if $(a-b)(a-c) < 0$?

Vrangr wrote: @above \[\frac{(a-b)(a-c)}{2a^2 + (b+c)^2} \ge \frac{(a-b)(a-c)}{2(a^2+b^2+c^2)}\]Is not always true, so your proof is not correct (what I mean is that this fact does not trivialize the problem). See post #3 for a proof using $2(b^2+c^2) \ge (b+c)^2$. Why it's not always true?

Zhero wrote: Let $a,b,c$ be positive reals. Prove that \[ \frac{(a-b)(a-c)}{2a^2 + (b+c)^2} + \frac{(b-c)(b-a)}{2b^2 + (c+a)^2} + \frac{(c-a)(c-b)}{2c^2 + (a+b)^2} \geq 0. \] Calvin Deng. Clearing the denominator, we get $\sum_{cyc}a^6+\left(\sum_{cyc}a^5b+\sum_{cyc}a^5c\right)+5\sum_{cyc}a^4bc+7\sum_{cyc}a^3b^3+15a^2b^2c^2\geq \left(\sum_{cyc}a^4b^2+\sum_{cyc}a^4c^2\right)+9\left(\sum_{cyc}a^3b^2c+\sum_{cyc}a^3bc^2\right)$ I couldn't solve this inequality.

Good try @Takeya.O .

Takeya.O wrote: Zhero wrote: Let $a,b,c$ be positive reals. Prove that \[ \frac{(a-b)(a-c)}{2a^2 + (b+c)^2} + \frac{(b-c)(b-a)}{2b^2 + (c+a)^2} + \frac{(c-a)(c-b)}{2c^2 + (a+b)^2} \geq 0. \] Calvin Deng. Clearing the denominator, we get $\sum_{cyc}a^6+\left(\sum_{cyc}a^5b+\sum_{cyc}a^5c\right)+5\sum_{cyc}a^4bc+7\sum_{cyc}a^3b^3+15a^2b^2c^2\geq \left(\sum_{cyc}a^4b^2+\sum_{cyc}a^4c^2\right)+9\left(\sum_{cyc}a^3b^2c+\sum_{cyc}a^3bc^2\right)$ I couldn't solve this inequality. When I saw it today, I solved easily. By Schur inequality, $5\sum_{cyc}a^4bc+15a^2b^2c^2\geq 5\left(\sum_{cyc}a^3b^2c+\sum_{cyc}a^3bc^2\right)$ By Muirhead inequality, $7\sum_{cyc}a^3b^3\geq \frac{7}{2}\left(\sum_{cyc}a^3b^2c+\sum_{cyc}a^3bc^2\right)$ and $\sum_{cyc}a^5b+\sum_{cyc}a^5c\geq \sum_{cyc}a^4b^2+\sum_{cyc}a^4c^2$ and $\sum_{cyc}a^6\geq \frac{1}{2}\left(\sum_{cyc}a^3b^2c+\sum_{cyc}a^3bc^2\right)$ So we're done.

So after fighting this through homogenization and Jensen, and through $uvw$ I think I got it equivalent to $108w^6 +72w^3v^2 +2w^3 +54v^6 -486v^8-81w^6v^2 -3v^2 +729w^6v^6 +486w^3v^6 +27v^4 \ge 0$

Hopefully this works. If so, this inequality is surprisingly weak! Observe that: $$2\cdot \frac{(a-b)(a-c)}{2a^2+(b+c)^2}=1-\frac{b^2+c^2+2a(b+c)}{2a^2+(b+c)^2},$$where similar identities hold for the other two terms. Hence after multiplying the given inequality by two, substituting, and squaring, the inequality is equivalent to: $$\left(\sum_{\text{cyc}} \frac{b^2+c^2+2a(b+c)}{2a^2+(b+c)^2}\right)^2\leq 9.$$From Cauchy-Schwarz, we have: $$\left(\sum_{\text{cyc}} \frac{b^2+c^2+2a(b+c)}{2a^2+(b+c)^2}\right)^2 \leq \left(\sum_{\text{cyc}} (b^2+c^2+2ab+2ac)\right)\left(\sum_{\text{cyc}} \frac{1}{2a^2+(b+c)^2}\right).$$Noting that $$\left(\sum_{\text{cyc}} (b^2+c^2+2ab+2ac)\right)=2(a+b+c)^2,$$set $a+b+c=3$. Then define: $$f(x)=\frac{1}{2x^2+(3-x)^2}=\frac{1}{3x^2-6x+9}.$$It's easy to see that it suffices to prove the stronger inequality: $$f(a)+f(b)+f(c)\leq \frac{1}{2},$$given $a+b+c=3$. Observe that $3x^2-6x+9$ is always positive, hence $f(x)$ is maximized when $3x^2-6x+9$ is minimized. It is easy to see that the minimum value of $3x^2-6x+9$ occurs when $x=1$, where it's value is $6$. Hence we have: $$f(x) \leq \frac{1}{6}~\forall x\in \mathbb{R}.$$Summing this over $a,b,c$ yields the desired inequality. $\blacksquare$

We use SOS Method. Since \[ \begin{aligned} \sum_{cyc} \frac{(a-b)(a-c)}{2a^2+(b+c)^2} &= \sum_{cyc} \frac{1}{2}\sum_{cyc}(b-c)(\frac{(b-a)}{2b^2+(a+c)^2}-\frac{(c-a)}{2c^2+(a+b)^2}) \\ &= \sum_{cyc} (b-c) \frac{(b-a)(a^2+b^2+2c^2+2ab) - (c-a)(a^2+2b^2+c^2+2ac)}{(2b^2+(a+c)^2)(2c^2+(a+b)^2)} \\ &= \sum_{cyc} (b-c)^2(\frac{-a^2+b^2+c^2-bc+3ab+3ac}{(2b^2+(a+c)^2)(2c^2+(a+b)^2)}) \end{aligned} \]Denote $S_A = \tfrac{-a^2+b^2+c^2-bc+3ab+3ac}{(2b^2+(a+c)^2)(2c^2+(a+b)^2)}$, $S_B, S_c$ defined symmetrically. WLOG, $a \ge b \ge c$, then $S_B, S_C > 0$. By SOS Theorem, it suffices to show that $a^2S_B+b^2S_A \ge 0$. And that can be done easily since \[ \begin{aligned} & \quad a^2(a^2+2b^2+c^2+2ac)(a^2-b^2+c^2-ac+3ab+3ac)+ \\ & \quad b^2(2a^2+b^2+c^2)(-a^2+b^2+c^2-bc+3ac+3ab) \\ &\ge a^2(a^2+2b^2+c^2+2ac)(a^2)-b^2(2a^2+b^2+c^2+2bc)a^2 \\ &= a^2[(a^4-b^4)+(a^2c^2-b^2c^2)+(2a^3c-2b^3c)] \ge 0 \end{aligned} \]Hence the original inequality is true, as desired.

We have \[ \frac{(a-b)(a-c)}{2a^2 + (b+c)^2} - \frac{2a^2-b^2-c^2-2a(b+c)+4bc}{4(a+b+c)^2} = \frac{[b^2+c^2+2a(b+c)](2a-b-c)^2}{4[2a^2+(b+c)^2](a+b+c)^2} \geqslant 0.\]Therefore \[\sum \frac{(a-b)(a-c)}{2a^2 + (b+c)^2} \geqslant \sum \frac{2a^2-b^2-c^2-2a(b+c)+4bc}{4(a+b+c)^2} = 0.\]

IAmTheHazard wrote: From Cauchy-Schwarz, we have: $$\left(\sum_{\text{cyc}} \frac{b^2+c^2+2a(b+c)}{2a^2+(b+c)^2}\right)^2 \leq \left(\sum_{\text{cyc}} (b^2+c^2+2ab+2ac)\right)\left(\sum_{\text{cyc}} \frac{1}{2a^2+(b+c)^2}\right).$$ I think the reason you find that the inequality is weak, is because I think this is not correct.